= Solution
For every $x^n\in A^{n,\varepsilon}$, the defining <typical set> constraint implies
$$
2^{-n(H(p)+\varepsilon)}\leq p_n(x^n)\leq2^{-n(H(p)-\varepsilon)}.
$$
Let $S_n=A^{n,\varepsilon}\cap B^{n,\varepsilon}$. Summing the lower <probability mass> bound over $S_n$ gives
$$
1\geq\Pr(X^n\in S_n)
=\sum_{x^n\in S_n}p_n(x^n)
\geq |S_n|2^{-n(H(p)+\varepsilon)}.
$$
Therefore
$$
\boxed{|A^{n,\varepsilon}\cap B^{n,\varepsilon}|\leq2^{n(H(p)+\varepsilon)}.}
$$
The <sample mean> constraint can only shrink the <typical set>, so this <cardinality> upper bound holds for every positive integer $n$.
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